1. Forces on each wire per unit length

Since the wires are long and parallel, each wire experiences:

• magnetic repulsion due to the other wire,
• weight due to its own mass,
• tension in the thread.

Let the mass per unit length of each wire be $\lambda$.

So, weight per unit length is $w=\lambda g.$

2. Magnetic force per unit length between the two wires

For two long parallel wires separated by distance $r$ and carrying current $I$ each, $\frac{F_m}{\ell}=\frac{\mu_0 I^2}{2\pi r}.$

3. Geometry of separation

Each thread has length $L$ and makes angle $\theta$ with the vertical. So each wire is displaced horizontally by $L\sin\theta.$

Hence separation between the two wires is $r=2L\sin\theta.$

Therefore magnetic force per unit length becomes $\frac{F_m}{\ell}=\frac{\mu_0 I^2}{2\pi(2L\sin\theta)}=\frac{\mu_0 I^2}{4\pi L\sin\theta}.$

4. Equilibrium condition

For equilibrium of each wire, resolve tension per unit length:

• horizontal component balances magnetic force,
• vertical component balances weight.

Thus, $T\sin\theta=\frac{F_m}{\ell},$ $T\cos\theta=\lambda g.$

Dividing, $\tan\theta=\frac{(F_m/\ell)}{\lambda g}.$

Substitute $F_m/\ell$: $\tan\theta=\frac{1}{\lambda g}\cdot \frac{\mu_0 I^2}{4\pi L\sin\theta}.$

So, $\lambda g\tan\theta=\frac{\mu_0 I^2}{4\pi L\sin\theta}.$

Using $\tan\theta=\frac{\sin\theta}{\cos\theta},$ we get $\lambda g\frac{\sin\theta}{\cos\theta}=\frac{\mu_0 I^2}{4\pi L\sin\theta}.$

Hence, $I^2=\frac{4\pi \lambda g L\sin^2\theta}{\mu_0\cos\theta}.$

Therefore, $I=2\sin\theta\sqrt{\frac{\pi \lambda g L}{\mu_0\cos\theta}}.$

5. Matching with options

This matches Option D: $2\sin \theta \sqrt {\frac{\pi \lambda gL}{{\mu _0}\,\cos \theta }}$

So the correct answer is D.